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\(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x-5\right)\left(x+5\right)=264\)
\(\Leftrightarrow x^3+64-x^3+25x=264\)
\(\Leftrightarrow25x=200\)
hay x=8
Đặt \(f\left(x\right)=x^3-2x^2-6x+a\)
Gọi thương của \(f\left(x\right):\left(x-2\right)\)là \(P\left(x\right)\)
\(\Rightarrow f\left(x\right)=P\left(x\right).\left(x-2\right)\)
Thay \(x=2\)ta có:
\(8-8-12+a=0\)
\(\Rightarrow a=12\)
Vậy \(a=2\)là giá trị cần tìm
\(a)6x^2y+9xy^2-2-3y=3xy\left(2x+3y\right)-\left(2x+3y\right)=\left(3xy-1\right)\left(2x+3y\right)\)
\(b)x^2-y^2+4-4x=\left(x-2\right)^2-y^2=\left(x+y-2\right)\left(x-y-2\right)\)
\(c)x^6-y^6=\left(x^3-y^3\right)\left(x^3+y^3\right)=\left(x-y\right)\left(x^2+y^2+xy\right)\left(x+y\right)\left(x^2+y^2-xy\right)\)
\(d)4x^2-9y^2+4x+1=\left(2x+1\right)^2-9y^2=\left(2x+3y-1\right)\left(2x-3y+1\right)\)
\(e)x^2-y^2+4x+4=\left(x+2\right)^2-y^2=\left(x+y+2\right)\left(x-y+2\right)\)
b,x2 -y2 +4-4x
=(x2 -4x +4)-y2
=(x-2)2 -y2
=(x-2-y)(x-2+y)
Để \(A=\frac{2x^2+3x+3}{2x+1}\)nguyên thì :
\(\left(2x^2+3x+3\right)⋮\left(2x+1\right)\)
\(\left(2x^2+x+2x+1+2\right)⋮\left(2x+1\right)\)
\(\left[x\left(2x+1\right)+\left(2x+1\right)+2\right]⋮\left(2x+1\right)\)
\(\left[\left(2x+1\right)\left(x+1\right)+2\right]⋮\left(2x+1\right)\)
Vì \(\left(2x+1\right)\left(x+1\right)⋮\left(2x+1\right)\)
\(\Rightarrow2⋮\left(2x+1\right)\)
\(\Rightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\Rightarrow x\in\left\{0;-1;0,5;-1,5\right\}\)
Vậy....
Ta có : 6x2 - 11x + 3
= 6x2 - 2x - 9x + 3
= (6x2 - 2x) - (9x - 3)
= 2x(3x - 1) - 3(3x - 1)
= (2x - 3)(3x - 1)
\(=\left(x^3-2x^2+x+2x^2-4x+2-2x+7\right):\left(x^2-2x+1\right)\\ =\left[\left(x^2-2x+1\right)\left(x+2\right)-2x+7\right]:\left(x^2-2x+1\right)\\ =x+2\left(dư:-2x+7\right)\)